## anonymous one year ago OMG!!!!!!!!!!! HELP PLZZZZZZZZ

1. anonymous

2. anonymous

@Sepeario

3. anonymous

@sweetburger

4. anonymous

@nincompoop

5. anonymous

@LunyMoony

6. anonymous

@imqwerty

7. anonymous

plzzzzz help

8. anonymous

hello??????????

9. Jack1

for this... u srsly need to flip a coin 50 times and fill in the table from ur results @bubbleslove1234 ... sorry, till u do we cant help u hey

10. anonymous

11. LunyMoony

yeah, I don't understand how they want the statistics either for the 2, or 2 each, of cats and dogs on the chart

12. Jack1

so what were ur answers for: HH - HT - TH - and TT? -

13. anonymous

hold on

14. anonymous

hh=11 ht=17 th=8 tt=14

15. Jack1

ok... so from ur screenshot: which question do u want to go through first?

16. anonymous

the first one and work from there

17. Jack1

k_ so cats are Heads, and Dogs are Tails according to the start of ur Q: having cat or dog is 50:50 "theoretical" probability so: Q1 - theoretical Probability of 2 cats OR 2 dogs = P(HH) OR P(TT) = P(HH) + P(TT) Theoretical probability of HH =0.25 Theoretical probability of TH =0.25 Theoretical probability of HT =0.25 Theoretical probability of TT =0.25 so solve = P(HH) + P(TT) @bubbleslove1234 ?

18. anonymous

do you know what i should get cause i am soooooooooooo confused

19. anonymous

oww my brain hurts

20. Jack1

Theoretical probability is: P(HH) =0.25 (or 25%) P(TH) =0.25 (or 25%) P(HT) =0.25 (or 25%) P(TT) =0.25 (or 25%) so Q1. solve = P(HH) + P(TT)

21. anonymous

okay so 25% times 50

22. Jack1

no, sorry u can either use percentages or use decimals, but u can't use both at the same time, sorry P(HH) = 0.25 P(TT) = 0.25 so = P(HH) + P(TT) =?

23. anonymous

okay

24. Jack1

P(HH) = 0.25 P(TT) = 0.25 so P(HH) + P(TT) = ????? @bubbleslove1234 ...?

25. anonymous

p(11)=0.25 p(14)=0.25

26. arindameducationusc

did you get the Question 1 answer? dog and cat?

27. anonymous

nope not yet but i need fast help cause it is due today

28. Jack1

no, for question 1, we only care about Theoretical results... not actual one we get from flipping a coin so Theoretical probability of 2 heads is 25% of the time... so P(HH) = 0.25 and the Theoretical probability of 2 tails is 25% of the time... so P(TT) = 0.25 so whats .25 + .25 ?

29. anonymous

50

30. Jack1

perfect. PHH + PTT = 0.50 .... or 50 % so Q1 = 50% chance

31. anonymous

oh so is that the answer

32. Jack1

Q1. chance family has 2 cats or 2 dogs? = P(CC) or P(DD) = 0.5 .... or 50% yep, answer

33. anonymous

for the first one

34. Jack1

Q2. how does 2 coins simulate which 2 pets a family has? answer: 50% chance of choosing a cat or a dog, and they chose twice (50% either cat or dog, then 50% chance either cat or dog for the 2nd pet also) so exactly as flipping a coin... twice (50% either heads or tail, then 50% chance either head or tail on the 2nd toss also)

35. Jack1

Q3.check if this is correct: ur results were: |dw:1439454960600:dw|

36. Jack1

|dw:1439455108667:dw|

37. Jack1

Q4. Experimental Probability In ur test, how many times did u flip the coin? @bubbleslove1234 ?

38. anonymous

50

39. Jack1

perfect now, calculate your results for P(HH) and P(TT) P(HH) = number of results that were HH / total number of flips =11/50 = what? (in decimals) P(TT) = number of results that were *TT / total number of flips =14/50 = what? (in decimals) @bubbleslove1234 ?

40. Jack1

sorry, typo

41. anonymous

P(HH) = 0.22 P(TT) =0.28

42. Jack1

absolutely right so: Experimental Probability of 2 cats or 2 dogs = same probability as flipping a coin 50 times and getting 2 heads or 2 tails so = P(HH) + P(TT) = 0.22 + 0.28 = ???

43. anonymous

0.5

44. anonymous

@Jack1

45. arindameducationusc

@bubbleslove1234 should I try?

46. anonymous

sure

47. arindameducationusc

okay for question 1, Dog/Cat 1 1 2 0 0 2 now, in the question its given 2 cats or 2 dogs, so trials are 3, so, its 2/3

48. Jack1

sorry my page froze... and is being glitchy can u finish @arindameducationusc please?

49. arindameducationusc

okay @Jack1 let me try.........

50. arindameducationusc

For question 2, just assign head as dog and tail as cat

51. anonymous

kk

52. arindameducationusc

now, hh=11 ht=17 th=8 tt=14 if this your data..... then count head head+tail+tail (according to question4 2 dogs and 2 cats) so, 11+14=25 now for probability divide by the frequency, so answer is 25/50

53. anonymous

okay

54. anonymous

got it

55. anonymous

can we move to Q5

56. arindameducationusc

yes sure,

57. anonymous

thanks

58. arindameducationusc

for question 5, C=Cat, D=Dog CCD CDD DDD DCC DDC CCC CDC DCD now, count where CCC and DDD is there which is 2. 2/8 , where 8 is the frequesncy. 2/8 is the probability

59. Jack1

hey, back and working, but keep going, happy to watch Q5: perfect, but 1 question: how did u get 2/8 for Q5... but only 2/3 for Q1? should it not be 2/4...?

60. arindameducationusc

Q5 and Q1 has no relation (very important)

61. arindameducationusc

one sec....

62. Jack1

no. categorically: no. Q1 and Q5 are totally related ;)

63. arindameducationusc

Q5 is correct.....

64. arindameducationusc

but Q1 I missed something.......

65. anonymous

what

66. arindameducationusc

if D=Dog C=Cat CC DD CD DC so yes it is 2/4. Thank you @Jack1

67. Jack1

np dude ;)

68. arindameducationusc

Now question 6

69. anonymous

kk

70. arindameducationusc

H=Head, T=Tail In the table given in Question, replace that with TTH THH HHH HTT HHT TTT THT HTH

71. anonymous

???????

72. anonymous

what

73. Jack1

he means : flip a coin 3 times, and record the results current simulation is; flip a coin 2 times

74. anonymous

75. Jack1

76. arindameducationusc

@Jack1 Exactly ! Flipping coins 3 times.... and record observations

77. anonymous

only 3 times

78. Jack1

flip a coin... 3 times.... record the results... repeat 50 times or flip 3 coins at the same time... 50 times

79. anonymous

okay

80. anonymous

thanks

81. arindameducationusc

Good job @bubbleslove1234 ... Was my attempt useful @bubbleslove1234 ?

82. anonymous

yep